3.159 \(\int \cos ^3(a+b x) \csc ^4(2 a+2 b x) \, dx\)

Optimal. Leaf size=43 \[ -\frac {\csc ^3(a+b x)}{48 b}-\frac {\csc (a+b x)}{16 b}+\frac {\tanh ^{-1}(\sin (a+b x))}{16 b} \]

[Out]

1/16*arctanh(sin(b*x+a))/b-1/16*csc(b*x+a)/b-1/48*csc(b*x+a)^3/b

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Rubi [A]  time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4287, 2621, 302, 207} \[ -\frac {\csc ^3(a+b x)}{48 b}-\frac {\csc (a+b x)}{16 b}+\frac {\tanh ^{-1}(\sin (a+b x))}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Csc[2*a + 2*b*x]^4,x]

[Out]

ArcTanh[Sin[a + b*x]]/(16*b) - Csc[a + b*x]/(16*b) - Csc[a + b*x]^3/(48*b)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 4287

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/e^p, Int[(e*Cos
[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rubi steps

\begin {align*} \int \cos ^3(a+b x) \csc ^4(2 a+2 b x) \, dx &=\frac {1}{16} \int \csc ^4(a+b x) \sec (a+b x) \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=-\frac {\operatorname {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=-\frac {\csc (a+b x)}{16 b}-\frac {\csc ^3(a+b x)}{48 b}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=\frac {\tanh ^{-1}(\sin (a+b x))}{16 b}-\frac {\csc (a+b x)}{16 b}-\frac {\csc ^3(a+b x)}{48 b}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 31, normalized size = 0.72 \[ -\frac {\csc ^3(a+b x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\sin ^2(a+b x)\right )}{48 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Csc[2*a + 2*b*x]^4,x]

[Out]

-1/48*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Sin[a + b*x]^2])/b

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fricas [B]  time = 0.45, size = 94, normalized size = 2.19 \[ \frac {3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 6 \, \cos \left (b x + a\right )^{2} + 8}{96 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

1/96*(3*(cos(b*x + a)^2 - 1)*log(sin(b*x + a) + 1)*sin(b*x + a) - 3*(cos(b*x + a)^2 - 1)*log(-sin(b*x + a) + 1
)*sin(b*x + a) - 6*cos(b*x + a)^2 + 8)/((b*cos(b*x + a)^2 - b)*sin(b*x + a))

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giac [A]  time = 0.93, size = 52, normalized size = 1.21 \[ -\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{96 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

-1/96*(2*(3*sin(b*x + a)^2 + 1)/sin(b*x + a)^3 - 3*log(abs(sin(b*x + a) + 1)) + 3*log(abs(sin(b*x + a) - 1)))/
b

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maple [A]  time = 1.21, size = 47, normalized size = 1.09 \[ -\frac {1}{48 b \sin \left (b x +a \right )^{3}}-\frac {1}{16 b \sin \left (b x +a \right )}+\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^4,x)

[Out]

-1/48/b/sin(b*x+a)^3-1/16/b/sin(b*x+a)+1/16/b*ln(sec(b*x+a)+tan(b*x+a))

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maxima [B]  time = 0.66, size = 834, normalized size = 19.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

1/96*(4*(3*sin(5*b*x + 5*a) - 10*sin(3*b*x + 3*a) + 3*sin(b*x + a))*cos(6*b*x + 6*a) + 36*(sin(4*b*x + 4*a) -
sin(2*b*x + 2*a))*cos(5*b*x + 5*a) + 12*(10*sin(3*b*x + 3*a) - 3*sin(b*x + a))*cos(4*b*x + 4*a) + 3*(2*(3*cos(
4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*cos(6*b*x + 6*a) - cos(6*b*x + 6*a)^2 + 6*(3*cos(2*b*x + 2*a) - 1)*cos(
4*b*x + 4*a) - 9*cos(4*b*x + 4*a)^2 - 9*cos(2*b*x + 2*a)^2 + 6*(sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin(6*b*x
+ 6*a) - sin(6*b*x + 6*a)^2 - 9*sin(4*b*x + 4*a)^2 + 18*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - 9*sin(2*b*x + 2*a
)^2 + 6*cos(2*b*x + 2*a) - 1)*log((cos(b*x + 2*a)^2 + cos(a)^2 - 2*cos(a)*sin(b*x + 2*a) + sin(b*x + 2*a)^2 +
2*cos(b*x + 2*a)*sin(a) + sin(a)^2)/(cos(b*x + 2*a)^2 + cos(a)^2 + 2*cos(a)*sin(b*x + 2*a) + sin(b*x + 2*a)^2
- 2*cos(b*x + 2*a)*sin(a) + sin(a)^2)) - 4*(3*cos(5*b*x + 5*a) - 10*cos(3*b*x + 3*a) + 3*cos(b*x + a))*sin(6*b
*x + 6*a) - 12*(3*cos(4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*sin(5*b*x + 5*a) - 12*(10*cos(3*b*x + 3*a) - 3*co
s(b*x + a))*sin(4*b*x + 4*a) - 40*(3*cos(2*b*x + 2*a) - 1)*sin(3*b*x + 3*a) + 120*cos(3*b*x + 3*a)*sin(2*b*x +
2*a) - 36*cos(b*x + a)*sin(2*b*x + 2*a) + 36*cos(2*b*x + 2*a)*sin(b*x + a) - 12*sin(b*x + a))/(b*cos(6*b*x +
6*a)^2 + 9*b*cos(4*b*x + 4*a)^2 + 9*b*cos(2*b*x + 2*a)^2 + b*sin(6*b*x + 6*a)^2 + 9*b*sin(4*b*x + 4*a)^2 - 18*
b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 9*b*sin(2*b*x + 2*a)^2 - 2*(3*b*cos(4*b*x + 4*a) - 3*b*cos(2*b*x + 2*a)
+ b)*cos(6*b*x + 6*a) - 6*(3*b*cos(2*b*x + 2*a) - b)*cos(4*b*x + 4*a) - 6*b*cos(2*b*x + 2*a) - 6*(b*sin(4*b*x
+ 4*a) - b*sin(2*b*x + 2*a))*sin(6*b*x + 6*a) + b)

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mupad [B]  time = 0.06, size = 38, normalized size = 0.88 \[ \frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{16\,b}-\frac {\frac {{\sin \left (a+b\,x\right )}^2}{16}+\frac {1}{48}}{b\,{\sin \left (a+b\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3/sin(2*a + 2*b*x)^4,x)

[Out]

atanh(sin(a + b*x))/(16*b) - (sin(a + b*x)^2/16 + 1/48)/(b*sin(a + b*x)^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3/sin(2*b*x+2*a)**4,x)

[Out]

Timed out

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